Question

The number of values of $\alpha$ in $[0,2\pi ]$ for which $2si{n}^3\alpha -7si{n}^2\alpha +7sin\alpha =2$ is

• A. 6
• B. 4
• C. 3
• D. 1

Answer 1

Answer: (C)

3

$2{\sin }^3\alpha -7{\sin }^2\alpha +7\sin \alpha -2=0$

Clearly $\sin \alpha =1$ is a solution

By polynomial division, we get

$(\sin \alpha -1)(2{\sin }^2\alpha -5\sin \alpha +2)=0$

$\Rightarrow (\sin \alpha -1)(2{\sin }^2\alpha -4\sin \alpha -\sin \alpha +1)=0$

$\Rightarrow (\sin \alpha -1)\left(2\sin \alpha \right(\sin \alpha -2)-(\sin \alpha -2\left)\right)=0$

$\Rightarrow (\sin \alpha -1)(2\sin \alpha -1)(\sin \alpha -2)=0$

$\Rightarrow \sin \alpha =1$ or $\sin \alpha =\frac{1}{2}$ or $\sin \alpha =2$

(not possible).

$\alpha =\frac{\pi }{2}$ or $\alpha =\frac{\pi }{6}$ or $\frac{5\pi }{6}$

$\therefore$ No. of solutions for $\alpha \in [0,2\pi ]$ is $3$.